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BITSPilaniHyderabad Campus

Dr.R.GururajCS&IS Dept.

Database Design & Applications

(SS ZG 518)

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BITS Pilani, Hyderabad Campus

Lecture Session-10

Indexing

Content

What is Indexing

Primary and Secondary indexes

Dense and Sparse Indexing

Multilevel Indexing

Designing Primary and Multilevel Indexes

What is Tree Indexing

B+ tree

Inserting and deleting keys into B+ Trees

Constructing a B+ tree

Designing a B+ Tree node structure

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An indexfor a file works in much the same way as a catalog

in a library.

In a library cards are kept in alphabetical order. So we dont

have to search all cards.

In real world databases, indexes may be too large to behandled efficiently.

Hence some sophisticated techniques are to be used.

Techniques for efficient retrieval of required records fromdisk are:

Hashing

Indexing

Introduction to Indexing


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The criteria for evaluating the hashing or indexing techniquesAccess time

Insertion time (new indexes or new records)

Deletion time

space overhead

Some times more than one indexing may be required for a file.

The attribute /field used for constructing index structure for a

file is called a indexing field/attribute .


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If the index field is a key, it is called as search key or indexing key.

Indexes on key attributes:

1. Built on ordering key(PK)Primary index

2. Non-ordering Key - Secondary index on key attribute

Indexes on non-key attributes:

1. Ordering non-key -- Clustering Index

2. Non-ordering non-key attributeSecondary index on non-key

Hence, a file can have at most one primary index or one clustering

index, but not both.


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Indexing

Nonordering field(secondary index)

Non-keykeyNon-key

(Clustering index)

Key

(primary index)

Ordering field


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Index record:Like data records, index records are alsostored in database. Any index record normally has two

fields.

Value Pointer

Key value Location address of

the record containing

the key

Data record:Similar kind of records(of a relation/table) arestored in a single file containing blocks. These are called

data records and will have fields specified on the relation.


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Dense Index: In this, an index record appears for every data file record.

Sparse Ind ex : Index records are created only for some data file

records. This occupies less space. Sparse index can be on primary or

secondary key.

A primary index and clustering index are non-dense.

SQL Commands to create indexes:

Usually when we declare PK, an index is created automatically.

CREATE INDEX EMP_IND ON EMP(eid);

DROP INDEX EMP_IND;


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2

15

25

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45

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Key

Index

Files/Blocks

2

5

6

9

Data files / Blocks

Pointer to

block

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35

6

9

Primary Indexing


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Ex 1:

Assume that we have an ordered file with 80000 recordsstored on disk. Block size is 512 Bytes. Record length isfixed and it is 70 Bytes. Key field(PK) length is 6 Bytesand block pointer is 4 Bytes. Assume unspanned recordorganization

Design a Primary index on primary key.

Design ing a Primary index


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Solution:

Sizeof disk block=512 Bytes; record length=70 Bytes

Block pointer=4 Bytes. Key field=6 bytes; total records=80000

No. records per block(Bfr)= floor (512/70)=7.31=7No. of data blocks needed= ceil(80000/7)= 11429

Index record length= key + pointer=6+4=10 Bytes

Blocking factor for index (Bfri) = floor(512/10)=51

(known as fanout)No. of index blocks = Ceil(11429/51)= 225

No. of block accesses= ceil of (log2 225) + 1 = 8+1=9


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KeyPointer tonext level

2

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Key

First Level

Second Level

2

5

6

9

Data files / Blocks

Pointer toblock

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Mult i level Ind exing (Two levels)


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Ex 2:

Assume that we have an ordered file with 80000 recordsstored on disk. Block size is 512 Bytes. Record length isfixed and it is 70 Bytes. Key field(PK) length is 6 Bytesand block pointer is 4 Bytes. Assume unspanned recordorganization

Design a multilevel index on primary key.

How many levels are there.

How many blocks are there in each index level.

Design ing a mu lt i level index


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Solution :

Size of the disk block=512 Bytes; record length=70 Bytes

Block pointer=4 Bytes. Key field=6 bytes; total records=80000

No. records per block(Bfr)= floor (512/70)=7.31=7

No. of data blocks needed= ceil(80000/7)= 11429Index record length= key + pointer=6+4=10 Bytes

Blocking factor for index = floor(512/10)=51 - fanout

No. of index blocks in first level= Ceil(311429/51)= 225

No. of index blocks in 2nd

level= Ceil(225/51)= 5No. of index blocks in 3rd level= Ceil(5/51)= 1 top level

No. of levels in indexing structure=t=3

No. of block accesses= No. index levels + 1= t+1=4


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B+ Treeis a multilevel search tree used to implementdynamic multilevel indexing. The primary disadvantage of

implementing multilevel indexes is that the performance

degrades as the file grows. It can be remedied by

reorganization, but frequent reorganization is not advisable.

B+ tree is best suited for multilevel indexing of files, because

it is dynamic.

B+ Tree of Orderp

It is a balanced tree, (all leaves are at same level).Each internal node is of the form-

B+ Tree Indexing

24

K1

32

K2

40

K3

60

K4

Child 1 Child 2 Child 3 Child 4Child 5


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Note

In a B+ tree record pointer for a record with given

key can be found only at leaf node.

But if it is in case of B-tree it can happen atintermediate node also.

Hence in B+ tree search, success or failure can be

declared only after reaching leaf_level.

Where as in B-tree search can be successful atintermediate level as well.

On failure we reach the leaf level.


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Constructing a B+ Tree

Ex 3:

Construct a B+ tree with given specifications. The order of the tree,

p=3 and pleaf=2. The tree should be such that all the keys in the

subtree pointed by a pointer which is preceding the key must be

equal to or less than the key value , and all the keys in the subtree

pointed by a pointer which is succeeding the key must be greater

than the key.

Insert the following keys in same order- 56, 22, 78, 42, 102, 90, 96,

35. Show how the tree will expand after each insertion, and the final

tree.

Next, delete 56, 46, 22 in the same order and show the status of

the tree after each deletion.


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Node design for B+ tree

Ex 5:We need to design a B+ tree indexing for Student

relation, on student_id attribute; the key of the relation.

The attribute student_id is of 4 bytes length. Other

attributes are- student_age(4 bytes), student_name(20bytes), student_address(40 bytes), student_branch(3

bytes). The Disk block size is 1024 Bytes. If the tree-

pointer takes 4 bytes, for the above situation, design the

best possible number of pointers per node(internal) of the

above B+ tree. Each internal node is a disk block which

contains search key values and pointers to subtrees.


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Solution:Disk block size=1024 Bytes

Size of B+ tree node= size of disk block

Each tree pointer points to disk block and takes 4 Bytes.

Each key (student_id) takes 4 Bytes

In a B+ tree node, No. of pointers = no. keys +1

Assume that no. keys = n

Then no. pointers= n+1

Then min. size for a node= {(no.Keys* size of each key)+

(no.pointers * size of each pointer)}

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Ex 6:In a data file which is ordered on the key field, we have 2,38,000 records.

The record length is 140 bytes and the block size is 1024 bytes. The

address of a disk block needs 8 bytes, and the key attribute of the file is of

9 bytes length.

If no indexing is done, give the number of block accesses needed (on

average) to retrieve a record with given key value from the above file. Also

give number of data blocks needed.

Now, Design a primary index for the above file on the key attribute. Give

how many index blocks are needed, and give the number of block

accesses needed (on average) to retrieve a record with given key value

from the above file.Now, Design a multilevel indexing for the same file and give, number of

levels with number of blocks at each level, and the number of block

accesses needed (on average) to retrieve a record with given key value

from the above file.

[Note: Complete working is required for your answer]24 10/09/2013 SSZG 518 Database Design & Applications Dr.R.Gururaj

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Given data :

Total No.of records = 2,38,000 records

Record length = 140 bytes

Block size = 1024 bytes

Key size = 9 bytes

Block pointer size = 8 bytes

Without Indexing:

Blocking factor for records of a file (Bfr) = floor(1024/140) = 7 records /

block

Total no. of data blocks required = Ceil (2,38,000/ 7) = 34,000

blocks/fileNo. of block access (on average) to access = ceil(log2 34000) = 16


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With primary indexingIndex entry size = 9 +8 = 17 bytes

Blocking factor for index = floor(1024/17) = 60

entries/block

Total no. of index blocks required = ceil(34,000/60)= 567

No. of block accesses needed to access = ceil(log2567) + 1= 10 +1 = 11

Data record (on average)

With Multi level indexing:

Blocking factor for index ( f0) = 60 entries / block

B1 (No.of index blocks) = 567

B2 (Second level index) = b1 / f0 = ceil(567 / 60) = 10 blocksB3 (Third level index) = b2 / f0 = ceil(10 / 60) = 1 block

Total no. of block accesses to data record = 3 + 1 = 4 block accesses.


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Summary

What is Indexing and its importance

How Primary and Secondary indexes work

Examples of Dense and Sparse Indexes

What is Multilevel Indexing

Some example problems on designing Primary

and Multilevel Indexes

What is Tree Indexing

B tree and B+ tree concepts

Constructing a B+ tree (Insert/Delete operations)

Designing a B+ Tree node structure

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